The load is 17156 N.
<u>Explanation:</u>
First compute the flexural strength from:
σ = FL / π
= 3000
(40
10^-3) / π (5
10^-3)^3
σ = 305
10^6 N / m^2.
We can now determine the load using:
F = 2σd^3 / 3L
= 2(305
10^6) (15
10^-3)^3 / 3(40
10^-3)
F = 17156 N.
Overcurrent protective devices, or OCPDs
Answer:
a. 0.4544 N
b. 
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction



= 0.27264 g


= 0.006816 mol
Now
Moles of
needed is

= 0.003408 mol


= 0.333984 g
Now based on the above calculation
a. Normality of acid is


= 0.4544 N
b. And, the acid solution molarity is


= 0.00005112
=
We simply applied the above formulas
Answer:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
Answer: Option D is not true of hydraulic valves. A hydraulic valve is a device that can change the opening degree of liquid flow path
Explanation:
The pilot check valve allows flow of liquid in one direction and blocks flow in the opposite direction